When monochromatic light of wavelength 620 nm is used to illuminate a metallic surface, the maximum kinetic energy of photoelectrons emitted is 1 eV. Find the maximum kinetic energy of photoelectron emitted (in eV) if a wavelength of 155 nm is used on the same metallic surface.
(hc = 1240 eV-nm)

Answer

To solve the problem of finding the maximum kinetic energy of photoelectrons emitted when monochromatic light of wavelength 155 nm is used, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of sufficient energy strikes a metallic surface, it can eject electrons from that surface. The energy of the incoming photons is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength of the light.

### Step 2: Calculate the Work Function (\( \Phi \))
From the first part of the question, we know that when light of wavelength 620 nm is used, the maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons is 1 eV. Using the photoelectric equation:
\[ E = \Phi + KE_{max} \]
we can rearrange it to find the work function:
\[ \Phi = E - KE_{max} \]

### Step 3: Calculate the Energy of the Photon at 620 nm
Using the formula for energy:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
- \( h \cdot c = 1240 \) eV·nm,
- \( \lambda = 620 \) nm.

Thus,
\[ E = \frac{1240 \text{ eV·nm}}{620 \text{ nm}} = 2 \text{ eV} \]

### Step 4: Find the Work Function
Now, substituting the values into the work function equation:
\[ \Phi = 2 \text{ eV} - 1 \text{ eV} = 1 \text{ eV} \]

### Step 5: Calculate the Energy of the Photon at 155 nm
Now we use the same formula for the new wavelength of 155 nm:
\[ E' = \frac{hc}{\lambda'} \]
Substituting the values:
\[ E' = \frac{1240 \text{ eV·nm}}{155 \text{ nm}} \approx 8 \text{ eV} \]

### Step 6: Calculate the Maximum Kinetic Energy for 155 nm
Using the photoelectric equation again:
\[ KE_{max}' = E' - \Phi \]
Substituting the values:
\[ KE_{max}' = 8 \text{ eV} - 1 \text{ eV} = 7 \text{ eV} \]

### Final Answer
The maximum kinetic energy of the photoelectrons emitted when light of wavelength 155 nm is used is:
\[ \boxed{7 \text{ eV}} \]